Question about the current flow and voltages in a simple NAND diagram

I've so far learned about boolean algebra and its applications to transistors. This is a picture of a NAND diagram:

A NAND has this boolean property : $f(A,B) = \overline{A \cap B}$. What I want to know is how the current is flowing (increasing/decreasing) and the voltages across the collectors. For example, lets say a voltage is applied to A but not to B, then the output which is a load lights up. From my understanding, you need a high enough voltage applied to the base of of a transistor to turn it on and once the base is saturated, you'll have enough current flowing from the collector to emitter. The current on the collector decreases which should mean a low voltage reading on the collector, but this transistor is connected to another transistor. Why is B turned off creating this reading across the resistor?

• Is point B floating, or connected to 0V?
– user16324
Nov 29, 2014 at 10:53

2 Answers

If B is turned off then its collector is effectively open circuit and there is nowhere for A's Base-Emitter current to go, so it cannot turn on.

With both A and B turned off there is no current flow through the resistor, so output voltage equals Vcc.

The resistor and the two transistors form a voltage divider. A transistor that is completely off has a resistance in the megohms to gigaohms, so the resultant voltage is very close to the supply regardless of the state of the first transistor.